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Standard X

Mathematics

Angles in the Same Arc Are Equal

In the figure...

Question

In the figure, the chord AB of the circle with centre O, is produced to C such that BC = OB. CO is joined and produced to meet the circle in D. If ∠ACD = y and ∠AOD = x, show that x = 3y.

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Solution

In $△$ OBC, OB = BC
⇒ ∠BOC = ∠BCO = y (angles opposite to equal sides are equal)

∠OBA is the exterior angle of ∆BOC
So, ∠ABO = 2y (exterior angle is equal to the sum of interior opposite angles]
Similarly, ∠AOD is the exterior angle of ∆AOC
∴ x = 2y + y = 3y.

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In the figure, the chord AB of the circle with centre O, is produced to C such that BC = OB. CO is joined and produced to meet the circle in D. If ∠ACD = y and ∠AOD = x, show that x = 3y.

In △ OBC, OB = BC ⇒ ∠BOC = ∠BCO = y (angles opposite to equal sides are equal) ∠OBA is the exterior angle of ∆BOC So, ∠ABO = 2y (exterior angle is equal to the sum of interior opposite angles] Similarly, ∠AOD is the exterior angle of ∆AOC ∴ x = 2y + y = 3y.

In $△$ OBC, OB = BC
⇒ ∠BOC = ∠BCO = y (angles opposite to equal sides are equal)

∠OBA is the exterior angle of ∆BOC
So, ∠ABO = 2y (exterior angle is equal to the sum of interior opposite angles]
Similarly, ∠AOD is the exterior angle of ∆AOC
∴ x = 2y + y = 3y.